The number of ordered pairs (r, k) for which | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation: $6 \cdot ^{35} C_{r} = (k^2 - 3) \cdot ^{36} C_{r+1}$.Using the identity $^{n} C_{r} = \frac{n}{r} \cdot ^{n-1} C_{r-1}$ or sp

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given the equation: $6 \cdot ^{35} C_{r} = (k^2 - 3) \cdot ^{36} C_{r+1}$.Using the identity $^{n} C_{r} = \frac{n}{r} \cdot ^{n-1} C_{r-1}$ or specifically $^{n+1} C_{r+1} = \frac{n+1}{r+1} \cdot ^{n} C_{r}$,we can rewrite the equation:$k^2 - 3 = \frac{6 \cdot ^{35} C_{r}}{^{36} C_{r+1}} = \frac{6 \cdot ^{35} C_{r}}{\frac{36}{r+1} \cdot ^{35} C_{r}} = \frac{6(r+1)}{36} = \frac{r+1}{6}$.Since $k$ is an integer,$k^2 - 3$ must be a non-negative integer. Also,for the combinations to be defined,$0 \le r \le 35$.Thus,$k^2 = \frac{r+1}{6} + 3 = \frac{r+19}{6}$.For $k^2$ to be a perfect square,we test values of $r \in \{0, 1, ..., 35\}$:If $r=5$,$k^2 = \frac{5+19}{6} = 4 \Rightarrow k = \pm 2$.If $r=35$,$k^2 = \frac{35+19}{6} = 9 \Rightarrow k = \pm 3$.The ordered pairs $(r, k)$ are $(5, 2), (5, -2), (35, 3), (35, -3)$.There are $4$ such ordered pairs.

The number of ordered pairs $(r, k)$ for which $6 \cdot ^{35} C_{r} = (k^2 - 3) \cdot ^{36} C_{r+1}$,where $k$ is an integer,is:

Similar Questions

The number of ways in which an arrangement of $4$ letters of the word $PROPORTION$ can be made is

How many five-digit numbers can be formed using the digits $2, 0, 4, 3, 8$ if repetition of digits is not allowed?

The exponent of $3$ in $100!$ is

$^nC_r + ^{n-1}C_r + ...... + ^rC_r$ =

There are $5$ apples,$4$ mangoes,$3$ oranges,and $1$ each of $2$ other varieties of fruits. The number of ways of selecting at least one fruit of each kind is

Vedclass Test Series

Exam Paper Generator

Online Exam Module